Completed Binary Search - 2

First:Last_Position_SortedArray.py

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+# Approach:
+# We perform binary search twice: first to find the leftmost index (first occurrence) 
+# and second to find the rightmost index (last occurrence) of the target. 
+# If the target is not present, both searches will return -1. This approach guarantees 
+# O(log n) time complexity by utilizing binary search instead of linear traversal.
+
+# Time Complexity: O(log n)
+# Space Complexity: O(1)
+# Did this code successfully run on Leetcode: Yes
+# Any problem you faced while coding this: No
+
+from typing import List
+
+class Solution:
+    def searchRange(self, nums: List[int], target: int) -> List[int]:
+        def findLeft(nums, target):
+            left, right = 0, len(nums) - 1
+            while left <= right:
+                mid = (left + right) // 2
+                if nums[mid] < target:
+                    left = mid + 1
+                else:
+                    right = mid - 1  # Continue searching left part
+            return left
+
+        def findRight(nums, target):
+            left, right = 0, len(nums) - 1
+            while left <= right:
+                mid = (left + right) // 2
+                if nums[mid] <= target:
+                    left = mid + 1  # Continue searching right part
+                else:
+                    right = mid - 1
+            return right
+
+        # Find the leftmost and rightmost positions
+        left_index = findLeft(nums, target)
+        right_index = findRight(nums, target)
+
+        # Check if the target is not found
+        if left_index <= right_index and right_index < len(nums) and nums[left_index] == target:
+            return [left_index, right_index]
+        else:
+            return [-1, -1]

Minimum.py

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+# Approach:
+# We use binary search to find the minimum element in a rotated sorted array. 
+# The key is to compare the middle element with the right boundary; if mid > right, 
+# the minimum is in the right half; otherwise, it's in the left half. 
+# This allows us to locate the minimum efficiently in O(log n) time.
+
+# Time Complexity: O(log n)
+# Space Complexity: O(1)
+# Did this code successfully run on Leetcode: Yes
+# Any problem you faced while coding this: No
+
+from typing import List
+
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        left, right = 0, len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            # If mid element is greater than the right element, the min is in the right half.
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            # If mid element is smaller or equal to the right, min is in the left half (including mid).
+            else:
+                right = mid
+
+        # At the end of the loop, left == right pointing to the minimum element.
+        return nums[left]

Peak_element.py

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+# Approach:
+# We use a binary search to find a peak element. At each step, we compare the middle element with its right neighbor; 
+# if the middle element is smaller, a peak must exist on the right half, otherwise, it exists on the left half. 
+# This approach guarantees an O(log n) runtime by dividing the search space in half at each step.
+
+# Time Complexity: O(log n)
+# Space Complexity: O(1)
+# Did this code successfully run on Leetcode: Yes
+# Any problem you faced while coding this: No
+
+from typing import List
+
+class Solution:
+    def findPeakElement(self, nums: List[int]) -> int:
+        left, right = 0, len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            # If the middle element is smaller than the next element, 
+            # a peak exists in the right half.
+            if nums[mid] < nums[mid + 1]:
+                left = mid + 1
+            # Otherwise, a peak exists in the left half (including mid).
+            else:
+                right = mid
+
+        # At the end of the loop, left == right and it points to a peak element.
+        return left